Timeline for If $\pi_\ast A$ is graded-commutative, then is $A_\ast$ a lax monoidal functor?

7 events

when toggle format	what		by	license	comment
Dec 3, 2022 at 2:30	comment	added	Tyler Lawson		In the case of Morava K-theory, we have only one generator to check ($v_n$) and its two images in $\pi_(K(n) \otimes K(n)^{op})$ are equal as a consequence of the formulas for the right unit for $BP_ BP$. (So, I believe, for genuinely interesting reasons!)
Dec 3, 2022 at 2:26	comment	added	Tyler Lawson		@TimCampion For K(n) you are fine. Another way to express this is that we are checking the composite $$A_* \otimes A_^{op} \to \pi_(A \otimes A^{op}) \to \pi_* end(A).$$
Dec 3, 2022 at 0:59	vote	accept	Tim Campion
Dec 3, 2022 at 0:59	comment	added	Tim Campion		Ok thanks! At least we see that the condition "$\pi_\ast(A)$ is graded-commutative" is not sufficient to ensure that $A_\ast$ is lax monoidal, even non-symmetrically. Which heightens the mystery for me -- is $K(n)_\ast$ lax monoidal when $p = 2$, and if so why?...
Dec 2, 2022 at 21:56	history	edited	Tyler Lawson	CC BY-SA 4.0	added 490 characters in body
Dec 2, 2022 at 19:55	history	edited	Tyler Lawson	CC BY-SA 4.0	added 11 characters in body
Dec 2, 2022 at 19:30	history	answered	Tyler Lawson	CC BY-SA 4.0